Suppose we have $(x^2+1)^2+4f(x)\geq 0$ for some $x$. Then there exists some $y_0$, such that $$y_0^2-(x^2+1)y_0-f(x)=0.$$ We may also assume that $y_0\ne 0$, because the roots of the above quadratic can't both be $0$. Plugging $P(x, y_0)$ in the equation we get $$y_0f(x)^2=0,$$ and because $y_0\ne 0$, we must have $f(x)=0$.
This directly implies $f(x)\leq 0$ for all $x$. Suppose the function does not positive roots. This means that we must have $f(x)<\frac{-(x^2+1)^2}{4}$ for all positive $x$. If we plug $P(x, 0)$ in the equation, where $x$ is positive, we get $$f(0)=f(-f(x))<-\frac{(f(x)^2+1)^2}{4}<-\frac{\left(-\left(\frac{(x^2+1)^2}{4}\right)^2+1\right)^2}{4},$$ where the first inequality follows from $-f(x)>0$ and $f(x)<\frac{-(x^2+1)^2}{4}$, whereas the second one follows from the fact that squaring the inequality mentioned above implies $$f(x)^2>\left(\frac{-(x^2+1)^2}{4}\right)^2.$$
Seeing as the $RHS$ of the inequality is not bounded from below, we have reached a contradiction, therefore we must have a positive root $a$. Plugging $P\left(x, \frac{a}{x^2+1}\right)$ in the equation we get $$0\geq f\left(\left(\frac{a}{x^2+1}\right)^2-f(x)\right)=\frac{a}{x^2+1}f(x)^2\geq 0,$$ so $\boxed{f\equiv 0}$, which indeed is a solution.
Aside from two simple observations, your answer after the EDIT makes perfect sense.
One is that the constant $ k $ can't be an arbitrary real number, and must be nonnegative (which you may have implicitly taken into account, but haven't asserted explicitly). That's because if $ k < 0 $, then $ - k \in \mathbb R ^ + $, and thus we should have $ Q ( - k ) - ( - k ) = k $, or equivalently $ Q ( - k ) = 0 $, which can't happen as by definition, codomain of $ Q $ is $ \mathbb R ^ + $. Knowing $ k \ge 0 $, your claim that $ f ( x ) = \frac 1 { x + k } $ is a solution, works perfectly well.
The other one is the comment by @Dylan, which adresses your claim about $ f $ being decreasing. As the comment states, you haven't used the fact that $ f $ is decreasing, but I want to stress that you could prove that fact by changing your argument just a little bit. At that point, you know that $ f ( x + y ) = \frac { f ( x ) } { f ( x ) + y } = \frac 1 { \frac 1 { f ( x ) } + y } $. As $ y > 0 $, you have $ \frac 1 { f ( x ) } + y > \frac 1 { f ( x ) } $, and then $ f ( x + y ) = \frac 1 { \frac 1 { f ( x ) } + y } < \frac 1 { \frac 1 { f ( x ) } } = f ( x ) $, which proves what is desired.
I'd like to add another way of thinking, which is essentially your own argument, but may be useful in my opinion. You could observe from the beginning that $ 1 - y f ( x + y ) = \frac 1 { 1 + y f ( x ) } > 0 $, which shows that $ f ( x + y ) < \frac 1 y $. This means that for every $ x , y \in \mathbb R ^ + $ with $ x < y $ we must have $ f ( y ) < \frac 1 x $, or equivalently for every $ y \in \mathbb R ^ + $, $ f ( y ) \le \frac 1 y $, which in turn shows that letting $ k _ y = \frac 1 { f ( y ) } - y $, we must have $ k _ y \ge 0 $. Then you could rewrite $ f ( x + y ) = \frac 1 { \frac 1 { f ( x ) } + y } $ as $ f ( x + y ) = \frac 1 { \left( \frac 1 { f ( x ) } - x \right) + ( x + y ) } $, which simply means that for every $ x , y \in \mathbb R ^ + $ with $ x < y $, we have $ f ( y ) = \frac 1 { y + k _ x } $. This means that for every $ x , y \in \mathbb R ^ + $, if we choose $ z $ so that $ z > \max ( x , y ) $, we must have $ \frac 1 { z + k _ x } = f ( z ) = \frac 1 { z + k _ y } $, which proves that $ k _ x = k _ y $, so we could simply use a single nonnegative constant $ k $, and we're done.
Best Answer
First of all, note that the problem can be stated in the following equivalent form.
It's straightforward to verify that the constant zero function, the identity function and the additive inversion function all satisfy the required condition. We prove that these are the only solutions.
Letting $ k = f ( 0 ) $ and putting $ x = y = 0 $ in \eqref{0}, we get $ f \bigl( - k - f ( k ) \bigr) = 0 $. Then, plugging $ x = - k - f ( k ) $ and $ y = 0 $ in \eqref{0} we can see that $ f \bigl( - f ( k ) \bigr) = 0 $. Now, set $ x = - f ( k ) $ in \eqref{0} and consider the equation once with $ y = - k - f ( k ) $ and another time with $ y = - f ( k ) $. Comparing the two results, you can conclude $ k ^ 2 f ( k ) = 0 $, which together with $ f \bigl( - f ( k ) \bigr) = 0 $ implies $ k = 0 $.
Putting $ y = 0 $ in \eqref{0} gives $$ f \bigl( - f ( x ) \bigr) ^ 3 = - x f ( x ) ^ 2 \tag 1 \label 1 $$ for all $ x \in \mathbb R $, while plugging $ x = 0 $ in \eqref{0} yields $$ f \left( - f ^ 2 ( y ) \right) ^ 3 = - y ^ 2 f ( y ) \tag 2 \label 2 $$ for all $ y \in \mathbb R $. Letting $ x = f ( y ) $ in \eqref{1}, we get $ f \left( - f ^ 2 ( y ) \right) ^ 3 = - f ( y ) f ^ 2 ( y ) ^ 2 $, which together with \eqref{2} shows that for any $ y \in \mathbb R \setminus \{ 0 \} $ with $ f ( y ) \ne 0 $ we have $ f ^ 2 ( y ) ^ 2 = y ^ 2 $. But also note that if $ y , f ( y ) \ne 0 $ and $ f ^ 2 ( y ) = - y $, then \eqref{2} gives $ f ( y ) \in \{ - y , y \} $; $ f ( y ) = y $ cannot happen since it implies $ f ^ 2 ( y ) = y $ (which contradicts $ f ^ 2 ( y ) = - y $ as $ y \ne 0 $), and $ f ( y ) = - y $ is not possible because it implies $ f ( - y ) = f ^ 2 ( y ) = - y $, which then substituting $ - y $ for $ y $ in \eqref{2} implies $ f ( y ) = y $ (contradicting $ f ( y ) = - y $ as $ y \ne 0 $). Therefore, the case $ f ^ 2 ( y ) = - y $ is ruled out, and we must have $$ f ( y ) \ne 0 \implies f ^ 2 ( y ) = y \tag 3 \label 3 $$ for all $ y \in \mathbb R \setminus \{ 0 \} $. Now, assume that there exists $ y _ 0 \in \mathbb R \setminus \{ 0 \} $ with $ f ( y _ 0 ) \ne 0 $. Putting $ y = y _ 0 $ in \eqref{2} and using \eqref{3} we have $ f ( - y _ 0 ) ^ 3 = - y _ 0 ^ 2 f ( y _ 0 ) $. Using this and \eqref{3} and setting $ y = y _ 0 $ in \eqref{0}, we can see that if $ f ( x ) = 0 $ for some $ x \in \mathbb R $, then we must have $ x = 0 $. Hence, using \eqref{3} we can conclude $ f ^ 2 ( y ) = y $ for all $ y \in \mathbb R $. Consequently, substituting $ f ( x ) $ for $ x $ in \eqref{0}, we get $$ f ( - x - y ) ^ 3 + 3 y ( x + y ) f ( x ) + x ^ 2 f ( x ) + y ^ 2 f ( y ) = 0 \tag 4 \label 4 $$ for all $ x , y \in \mathbb R $. Interchanging $ x $ and $ y $ in \eqref{4} and comparing with \eqref{4} itself, it's straightforward to see that $$ y f ( x ) = x f ( y ) $$ for all $ x , y \in \mathbb R $. In particular, this gives $ f ( x ) = f ( 1 ) x $ for all $ x \in \mathbb R $, and as $ f ^ 2 ( 1 ) = 1 $, we have $ f ( 1 ) \in \{ - 1 , 1 \} $. Therefore, we've proven that if $ f $ differs from the constant zero function, we must either have $ f ( x ) = x $ for all $ x \in \mathbb R $ or $ f ( x ) = - x $ for all $ x \in \mathbb R $, which proves what was claimed.